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Momentum + coefficient of restitution this is driving me nuts

  1. Feb 26, 2010 #1
    Momentum + coefficient of restitution this is driving me nuts!!!

    1. The problem statement, all variables and given/known data

    A Mass of 2kg travelling at 2m/s is struck by a mass of 4 kg travelling at 3m/s. If the coefficient of restitution for the two materials is 0.4 calculate their final velocities.

    2. Relevant equations

    M1U1+M2U2=M1V1+M2V2

    E=V2-V1/U2-U1


    3. The attempt at a solution

    From what i know, the best way to do it is use both equations and create two simultaneous equations?

    Is this correct? i seem to end up with V2=-1.2 however im pretty sure this is wrong.

    any help would be great.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 26, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi ar202! Welcome to PF! :wink:

    Show us your full calculations, and then we'll see what went wrong, and we'll know how to help! :smile:
     
  4. Feb 26, 2010 #3
    Re: Momentum + coefficient of restitution this is driving me nuts!!!

    Cheers dude :-)

    first bit...

    2*2+4*(-4)=2*V1+4*V2

    -12=2*V1+4*V2

    second bit...

    0.4*2-0.4*4=v2-v1

    2.4=v2-v1

    so then i have

    -12=2*V1+4*V2
    +
    4.8=2V2-2V1

    which gives

    -7.2=6V2...
     
  5. Feb 26, 2010 #4

    tiny-tim

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    Hi ar202! :smile:

    erm :redface: … you keep writing 4 instead of 3 …
    :cry:
     
  6. Feb 26, 2010 #5
    Re: Momentum + coefficient of restitution this is driving me nuts!!!

    haha so i have :p

    the answer is still wrong... any tips?
     
  7. Feb 26, 2010 #6

    tiny-tim

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    Have you checked that your signs for V1 and V2 are the right way round in both equations?

    (and btw, were they travelling in the same direction originally, or in opposite directions?)
     
  8. Feb 26, 2010 #7
    Re: Momentum + coefficient of restitution this is driving me nuts!!!

    it doesn't mention directions, just that it is struck. I've found the answers - V1=2.53 & V2=2.93

    still struggling to get to those answers.
     
  9. Feb 26, 2010 #8

    tiny-tim

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    Well, then they're obviously going in the same direction originally …

    relative speed before = 1, after = 0.4 :wink:
     
  10. Feb 26, 2010 #9
    Re: Momentum + coefficient of restitution this is driving me nuts!!!

    Ok so using the same method as before i can get V2=2.83 & V1=2.34

    am i going about it the right way...?
     
  11. Feb 26, 2010 #10

    tiny-tim

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    hmm … relative speed after = 0.49 … not close enough to 0.4 :redface:

    you must be making a mistake somewhere, but without seeing the details, I can't say where.
     
  12. Feb 26, 2010 #11
    Re: Momentum + coefficient of restitution this is driving me nuts!!!

    first part

    2*2+4*3=2*v1+4*v2

    0.4(3-2)=v2-v1 X 2 then i've added it to the first part...

    so 18=6*v2

    18/6=3 .... actually my answer was 3 for v2. I must be more tired than i thought :p
     
  13. Feb 26, 2010 #12

    tiny-tim

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    oooh, i see it now …
    … there's no "X 2" …

    the coefficient of restitution is just |u1 - u2|/|v1 - v2| :wink:
     
  14. Feb 26, 2010 #13
    Re: Momentum + coefficient of restitution this is driving me nuts!!!

    the 'X2' was the simultaneous equation part...
     
  15. Feb 27, 2010 #14

    tiny-tim

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    (just got up :zzz: …)

    oh, it multiplied the whole equation!

    ok, then your 18 should be 16.8 :wink:
     
  16. Feb 27, 2010 #15
    Re: Momentum + coefficient of restitution this is driving me nuts!!!


    ok i finally got there :rolleyes:

    got it down to 6v2=15.8

    15.8/6=2.53333333333

    thanks for your help dude. think i made it alot harder than it should have been!! :blushing:
     
  17. Feb 27, 2010 #16

    tiny-tim

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    you do know that 15.2/6=2.53333333333 ? :wink:
     
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